SOLUTION: find all critical points and global extremes of the functions on the given intervals. 1. f(x) = x3 – 3x + 5 on the entire real number line. 2. f(x) = x3 – 3x + 5 on [ –2, 1]

Algebra ->  Functions -> SOLUTION: find all critical points and global extremes of the functions on the given intervals. 1. f(x) = x3 – 3x + 5 on the entire real number line. 2. f(x) = x3 – 3x + 5 on [ –2, 1]      Log On


   

Question 996744: find all critical points and global extremes of the functions on the given intervals.
1. f(x) = x3 – 3x + 5 on the entire real number line.
2. f(x) = x3 – 3x + 5 on [ –2, 1]
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
find all critical points and global extremes of the functions on the given intervals.
1. f(x) = x3 – 3x + 5 on the entire real number line.
graph%28400%2C400%2C-10%2C10%2C-10%2C10%2Cx%5E3-3x%2B5%29
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f'(x) = 3x^2 - 3
Sove for "x"::
3x^2 - 3 = 0
3(x^2-1) = 0
x = 1 or x = -1
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f"(x) = 6x
f"(1) = 6 so (1,f(1)) is a local minimum at x = 1.
f'(-1) = -6 so (-1,f(-1)) is a local maximum at x = -1
No global max or min
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2. f(x) = x3 – 3x + 5 on [ –2, 1]
f(-1) = (-1)+3+5 = 7 (global max)
f(1) = 1-3+5 = 3 ( global min)
f(-2) = -8+6+5 = 3
f(1) = 3
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Cheers,
Stan H.
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